Question

A body of mass 60 kg has momentum of 3000 kg m s^-1 .
Calculatev:
(i) the kinetic energy,
(ii) the speed of the body.
PLEASE HELP ME ASAP!!!​

Answer 1

Answer:

K.E=p

2

/2m

p = 3000 kg m/sec and m = 60 kg

This is the relation obtained from two formulae,

p=m×v and K.E=

2

1

×m×v

2

Hence K.E=

2m

p

2

=

2×60

3000

2

J

K.E=75×10

3

J.

Explanation:

follow.........

Answer 2

(i) Solution, m = 60 kg

And momentum (p) = 3000 kg m s⁻¹

kinetic energy (K.E.) =?

(K.E) = p²/2m

(K.E) = (3000 kg m s⁻¹)² /2×60 kg

(K.E) = 9000000 kg m² s⁻² / 120

(K.E) = 75000 kg m² s⁻²

Or, (K.E) = 75000 J = 75× 10³ Joules

(ii) speed of the body (v),

Momentum ( P )= mv

v (speed) = P / m

v(speed) = 3000 kg m¹ s⁻¹ / 60 kg

v(speed) = 50 m¹ s⁻¹

Thanks

Maybe u can find speed of the body by one more formula ,

(K.E.) = 1/2 mv²

75×10³ kg m² s⁻² = 1/2 (60 kg) v²

75×10³m² s⁻² = 30 v²

v² = 75×10³ m² s⁻² /30

v² = 2500 m²/s²

v = √2500 m²/s²

v = 50 m/s

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